Question: $f(x) = \begin{cases} x+4 & \text{for} ~~~~x\gt{-1} \\ -\dfrac3x& \text{for} ~~~~ x \leq-1\end{cases}$ Evaluate the definite integral. $\int^1_{-3}f(x)\,dx = $ Choose 1 answer: Choose 1 answer: (Choice A) A $3\ln\left(\frac13\right) +\dfrac92$ (Choice B) B $\ln(\frac13) -1$ (Choice C) C $3\ln(3) +8$ (Choice D) D Undefined
Solution: Splitting up the definite integral Since we're working with a piecewise function, we need to split the definite integral up into two pieces: $\phantom{=} \int^1_{-3}f(x)\,dx$ $= \int^1_{-1}f(x)\,dx + \int^{-1}_{-3}f(x)\,dx~~~~~~$ [Why did we split at -1?] $= \int^1_{-1}(x+4)\,dx + \int^{-1}_{-3}\left(-\dfrac3x\right)\,dx ~~~~~~$ Evaluating each piece Next, let's evaluate each of these definite integrals one at a time. The first definite integral: $\begin{aligned} \int^1_{-1}(x+4)\,dx &=\left(\dfrac12x^2+4x\right)\Bigg|^1_{{-1}} \\\\ &= \left[\dfrac12\left({1} \right)^2 +4\cdot(1)\right] - \left[ \dfrac12\left({-1} \right)^2 +4\cdot({-1})\right] \\\\ &= \left[\dfrac92\right] -\left[-\dfrac72 \right] \\\\ &= {8}\end{aligned}$ The second definite integral: $\begin{aligned} \int^{-1}_{-3}\left(-\dfrac3x\right)\,dx &=-3\ln(|x|)\Bigg|^{-1}_{{-3}} \\\\ &=-\left[-3\ln(| {-1}|) \right] - \left[-3\ln (|{-3}|)\right] \\\\ &=-\left[-3\ln( {1}) \right] - \left[-3\ln ({3})\right] \\\\ &= \left[-3\cdot0\right] -\left[-3\ln(3) \right] \\\\ &= {3\ln(3)}\end{aligned}$ Putting the pieces together Now let's add these two pieces together to find the answer: $\phantom{=} \int^1_{-1}(x+4)\,dx + \int^{-1}_{-3}\left(-\dfrac3x\right)\,dx$ $ ={8} + {3\ln(3)} $ The answer $\int^3_{0}f(x)\,dx = 3\ln(3) +8$